joigus

AAMOF, relativity makes your intent to supersede F=dp/dt with the 'bold new physics' F=ma even more implausible, as p has an extra dependence on v which makes the connection between F (in relativity the 4-force) one step more logically --and calculationally-- removed from acceleration. It has terms that do not involve second-order time derivatives. One term is proportional to velocity, and the other is proportional to acceleration. So no.

Agreed. I was thinking more about a rocket orbiting a planet. So external forces would be conservative.

My answer would be a resounding 'no' based on reasons pretty much pointed out by other members. I have little much significant to add to what, eg, Markus has said. If anything, it strikes me as a sample of our most primitive instincts hijacking our reason under the guise of being a 'rational' solution. It's essentially what our traditional approach to garbage has been throughout centuries: Round it up and put it away, I don't wanna see it anywhere near me.

Ok. In the notes that I'm taking to in order to understand the problem better, I'm distinguishing Fe,ext, Fr,ext, and Fext=Fe,ext+ Fr,ext, which might seem a little bit overthinking it, but is not too bad an approach if one wants to make sure we're not missing anything.

Nice pic. I do remember having watched cousins of these guys in some documentary. Different colour perhaps, but general appearance very similar.

Absolutely. I'm looking at it from every perspective I know, and I can't see any reason why this shouldn't hold true. I can't see any way in which dp/dt is not valid here either. I think we agree on that. Am I right?

I'm checking up on everything I can. I get the same as you. If motion of respective CoM of both rocket and exhaust are collinear and in the absence of external fields. Of course, forces must be identified with dp/dt. What's tricky is the momentum of what and what force on what 'object.' I'm considering the exhaust as one big indistinct thing, although its CoM must move in a predictable way.

OK. I'm analising everything in terms of Newtonian mechanics. I need some time to react to @studiot and @sethoflagos, and see if we agree on reference systems and everything else. You can either ignore external fields --or consider the system in free space-- or not. I'll think about a situation that's as general as possible without the whole thing being a mess. Real rockets, of course, require many other considerations.

Yes. The rocket --never mind it being an open system-- has its own Lagrange equation. The beauty of Lagrange is you don't even have to think about forces. It's all in describing so-called configurations of the system. That is: How many variables do I need to know where it's at in its evolution? You don't even need to consider whether you are in an inertial system or not. Coordinates could be curvilinear for all you care. Lagrange's method takes care of everything else, including inertial forces.

This is not even dimensionally correct. Besides, m in both sides of the equation would simplify. Please, review your maths before you commit another badly wrong comment.

No. Fext stands for 'total external force' for both the rocket --with unburnt fuel-- and the exhaust. You didn't understand me. I will need some more time, as comments are piling up. I see no end to this...

That's fairly clear to me. You don't understand... Here (from page 8, where you apparently answered 'several times': And indeed, \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \] Take \( F_{\textrm{ext}}=0 \), and there's your super-special instance. I told you, didn't I? More from unbearable page 8... I told you conservation of momentum applies for overall rocket + exhaust in free space, and I told you how. I've highlighted it. Maybe I'm lucky enough that you happen to read it this time. Etc. PS: By fuel + rocket I mean 'exhaust + rocket.' Good day, sir.

Dear @martillo, there's only one person here who's just been adhering to his own opinion, gritting his teeth against all evidence and reasoning. The evidence being that these are the concepts that engineers apply on a regular basis. All of this is established science. Nothing more to say on my part, as you totally ignored my arguments. Other members seem to have come to similar conclusions.

Lee Smolin: "The trouble with physics..." Martillo: "...is F=dp/dt" "is F=ma"

I know. This is very close to coming full circle, or is already there. Sigh.

What system? There's your vagueness again. Which is included as a particular case in my analysis, of which you said nothing.

Absolutely non-sensical piece of WAG pseudo-reasoning. You obviously don't understand what gravity and mass --inertial or gravitational-- are. You obviously don't understand what electric charge is. You obviously don't understand what zero-point energy is. And, BTW, axioms never justify your logic. Axioms are initial assumptions, and logic is a given.

No. It (momentum of either rocket, fuel or both) only has to be conserved if there are no external forces on the system. In the more general case with an external force that I presented to you --in the presence of this external force acting on both the rocket and the exhaust-- neither momentum nor (mechanical) energy have to be conserved. You say "momentum" and "energy" as if they must mean something independent of what system you apply it to, or which level of detail you want to describe it. They don't. Rocket: Energy is not conserved Momentum is not conserved Exhaust: Energy is not conserved Momentum is not conserved Rocket + exhaust: Mechanical energy is not conserved Mechanical energy + chemical energy stored in fuel is conserved Momentum is not conserved if rocket + exhaust are in a gravitational field Momentum is conserved if fuel + rocket are in free space You see, when you burn fuel, some kind of potential energy that's stored in the molecular bonds gets converted into heat + kinetic energy for both rocket and exhaust. We call this energy Gibbs free energy (of fuel). So no, mechanical energy is not conserved in this case. If the rocket is stationary in free space at initial time, K.E = P.E. = 0 initially. Then it starts burning fuel and starts moving. Now P.E. is not cero, but K.E. is K.E. of exhaust + K.E. of rocket. So where did this energy come from?

I wasn't showing lack of concern. I forgot to write a sentence that I had in my mind. Namely: "which was every bit as horrible as a WW-nth, never mind the number of casualties." The Cold War was pretty horrific, if you ask me, or people who suffered it in Angola, Chile, Korea, Vietnam, and on, and on. You can add to that many indirect consequences, like Chernobyl, which IMO was a consequence of the Soviets beeing obsessed with getting ahead of the US.

LOL ...

I agree. I'm no historian or political analyst, but this looks very much like a CWII rather than a WWIII, because the nukes are still there. Technology has completely changed the rules, and it keeps doing so. This has led to a whole newfangled model of warfare and intelligence. You only have to look at the influence of such things as the Internet, drones, and what's in store with the possibilities that AI offers.

Not necessarily. Here's a sketch of how it would be done. 1) We don't care about the exhaust at all. Our system is the rocket; but it's an open system. To simplify, let's make it 1-degree of freedom, with coordinate of rocket = \( x \). \[ L=\frac{1}{2}m\dot{x}^{2}-V\left(x\right) \] Momentum of rocket is partial derivative of \( L \) with respect to generalised velocity. Just following our noses, and blindly believing in variational calculus and our guessed-at Lagrangian being good: \[ p_{x}=\frac{\partial L}{\partial\dot{x}}=m\dot{x} \] Euler-Lagrange equation: \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \] which is, of course, the rocket equation. 2) We care about the exhaust. Take all this with a pinch of salt, because I'm not 100% sure this is correct. It produces consistent equation for the rocket though. It's possible that the law of change of \( m \) with time had better be treated as a time-dependent constraint, rather than a coordinate. The exhaust is complicated to describe. But, as we don't care about it very much, we will only describe it with one generalised coordinate, \( X \) wich accounts for the CoM of the whole thing at time \( t \). Very important: The system configuration needs specifying three coordinates: \( x \), \( X \) and \( m\left( t \right) \). We will have 3 Euler-Lagrange equations. The Lagrangian should be, \[ L=\frac{1}{2}\left(M-m\right)\dot{X}^{2}+\frac{1}{2}m\dot{x}^{2}-V_{r}\left(x\right)-V_{e}\left(X\right) \] and \( V_{r}\left(x\right) \), \( V_{e}\left(X\right) \) the respective potential energies of rocket and exhaust CoM. We will also treat \( m \) as a generalised coordinate. The Euler-Lagrange equations are, \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow m\ddot{x}+\dot{m}\dot{x}=-V_{r}'\left(x\right) \] \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{X}}-\frac{\partial L}{\partial X}=0\Rightarrow\frac{d}{dt}\left[\left(M-m\right)\dot{X}\right]=-V_{e}'\left(X\right) \] \[ \frac{d}{dt}\frac{\partial L}{\partial\dot{m}}-\frac{\partial L}{\partial m}=0\Rightarrow V_{r}\left(x\right)=V_{e}\left(X\right) \] And those are your evolution equations, at least within this very simple model, and as you can see you get a generalised expression for the momentum of the rocket that accounts for the mass it's losing. My conclusion thus is that @Genady --and everybody else-- is right. As I told you, your trusty momentum can be generalised. This version is called 'canonical momentum associated to coordinate \( x \).' In the case of a particle in a magnetic field, it has a 'tail' that contains the vector potential. It doesn't have to conform to any previous ideas that you have about what it is. The only thing that spoils your Lagrangian methods is dissipation in those cases where you can't follow where the energy has gone. If you don't care about details, and are happy to describe only the CoM of the bulk of exhaust, you can get a good understanding of what's going on. Forgot to mention: \( M \) is a certain total initial mass of rocket when it's fully charged of fuel.

I've never interviewed a muon. And nobody is saying that. Please, read @studiot's answer and ponder what he' saying.

Oh, come on. As Leonard Susskind put it, you can make a pretty good living today out of criticising every new idea that comes up. Beating string theory is beating a dead horse. It sells books too. There are many other ideas besides string theory. What's his problem with Witten in particular? Do you know anything about string theory? Your technique for renormalising a divergent logarithmic integral is making up its numerical value. Do I really need to say more?

Now I realise I meant to emphasise "indirect", not "evidence."